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Contents 3 Homomorphisms, Ideals, and Quotients Ring Theory (part 3): Homomorphisms, Ideals, and Quotients (by Evan Dummit, 2018, v. 1.01) Contents 3 Homomorphisms, Ideals, and Quotients 1 3.1 Ring Isomorphisms and Homomorphisms . 1 3.1.1 Ring Isomorphisms . 1 3.1.2 Ring Homomorphisms . 4 3.2 Ideals and Quotient Rings . 7 3.2.1 Ideals . 8 3.2.2 Quotient Rings . 9 3.2.3 Homomorphisms and Quotient Rings . 11 3.3 Properties of Ideals . 13 3.3.1 The Isomorphism Theorems . 13 3.3.2 Generation of Ideals . 14 3.3.3 Maximal and Prime Ideals . 17 3.3.4 The Chinese Remainder Theorem . 20 3.4 Rings of Fractions . 21 3 Homomorphisms, Ideals, and Quotients In this chapter, we will examine some more intricate properties of general rings. We begin with a discussion of isomorphisms, which provide a way of identifying two rings whose structures are identical, and then examine the broader class of ring homomorphisms, which are the structure-preserving functions from one ring to another. Next, we study ideals and quotient rings, which provide the most general version of modular arithmetic in a ring, and which are fundamentally connected with ring homomorphisms. We close with a detailed study of the structure of ideals and quotients in commutative rings with 1. 3.1 Ring Isomorphisms and Homomorphisms • We begin our study with a discussion of structure-preserving maps between rings. 3.1.1 Ring Isomorphisms • We have encountered several examples of rings with very similar structures. • For example, consider the two rings R = Z=6Z and S = (Z=2Z) × (Z=3Z). ◦ Here are the addition and multiplication tables in R: + 0 1 2 3 4 5 · 0 1 2 3 4 5 0 0 1 2 3 4 5 0 0 0 0 0 0 0 1 1 2 3 4 5 0 1 0 1 2 3 4 5 2 2 3 4 5 0 1 2 0 2 4 0 2 4 3 3 4 5 0 1 2 3 0 3 0 3 0 3 4 4 5 0 1 2 3 4 0 4 2 0 4 2 5 5 0 1 2 3 4 5 0 5 4 3 2 1 1 ◦ Now compare those tables to the tables in S: + (0; 0) (1; 1) (0; 2) (1; 0) (0; 1) (1; 2) · (0; 0) (1; 1) (0; 2) (1; 0) (0; 1) (1; 2) (0; 0) (0; 0) (1; 1) (0; 2) (1; 0) (0; 1) (1; 2) (0; 0) (0; 0) (0; 0) (0; 0) (0; 0) (0; 0) (0; 0) (1; 1) (1; 1) (0; 2) (1; 0) (0; 1) (1; 2) (0; 0) (1; 1) (0; 0) (1; 1) (0; 2) (1; 0) (0; 1) (1; 2) (0; 2) (0; 2) (1; 0) (0; 1) (1; 2) (0; 0) (1; 1) (0; 2) (0; 0) (0; 2) (0; 1) (0; 0) (0; 2) (0; 1) (1; 0) (1; 0) (0; 1) (1; 2) (0; 0) (1; 1) (0; 2) (1; 0) (0; 0) (1; 0) (0; 0) (1; 0) (0; 0) (1; 0) (0; 1) (0; 1) (1; 2) (0; 0) (1; 1) (0; 2) (1; 0) (0; 1) (0; 0) (0; 1) (0; 2) (0; 0) (0; 1) (0; 2) (1; 2) (1; 2) (0; 0) (1; 1) (0; 2) (1; 0) (0; 1) (1; 2) (0; 0) (1; 2) (0; 1) (1; 0) (0; 2) (1; 1) ◦ Notice that these tables look quite similar (especially given the artful reordering of the labels of the elements in S). ◦ Indeed, if we relabel each entry n in the rst set of tables with the ordered pair corresponding to its reduction modulo 2 and 3 (so that 1 becomes (1; 1), 2 becomes (0; 2), and so forth) we will obtain the second set of tables! For another example, consider the rings and 2 . • R = (Z=2Z) × (Z=2Z) S = F2[x]=(x + x) ◦ Here are the addition and multiplication tables in R: + (0; 0) (1; 1) (1; 0) (0; 1) · (0; 0) (1; 1) (1; 0) (0; 1) (0; 0) (0; 0) (1; 1) (1; 0) (0; 1) (0; 0) (0; 0) (0; 0) (0; 0) (0; 0) (1; 1) (1; 1) (0; 0) (0; 1) (1; 0) (1; 1) (0; 0) (1; 1) (1; 0) (0; 1) (1; 0) (1; 0) (0; 1) (0; 0) (1; 1) (1; 0) (0; 0) (1; 0) (1; 0) (0; 0) (0; 1) (0; 1) (1; 0) (1; 1) (0; 0) (0; 1) (0; 0) (0; 1) (0; 0) (0; 1) ◦ Now compare those tables to the tables in S: + 0 1 x x + 1 · 0 1 x x + 1 0 0 1 x x + 1 0 0 0 0 0 1 1 0 x + 1 x 1 0 1 x x + 1 x x x + 1 0 1 x 0 x x 0 x + 1 x + 1 x 1 0 x + 1 0 x + 1 0 x + 1 ◦ Here, if we relabel (0; 0) as 0, (1; 1) as 1, (1; 0) as x, and (0; 1) as x + 1, the rst pair of tables becomes the second set of tables. a b • As a third example, consider the rings R = = fa+bi : a; b 2 g and S = 2 M ( ): a; b 2 . C R −b a 2×2 R R a b c d a − c b − d ◦ Notice that S is a subring of M ( ): we have − = and 2×2 R −b a −d c −(b − d) a − c a b c d ac − bd ad + bc · = . −b a −d c −(ad + bc) ac − bd ◦ Compare these to the addition and multiplication operations in C: (a + bi) − (c + di) = (a − c) + (b − d)i and (a + bi) · (c + di) = (ac − bd) + (ad + bc)i. a b ◦ Upon identifying the complex number a + bi with the matrix , we see that the ring structure −b a of S is precisely the same as the ring structure of C. • Let us formalize the central idea in the examples above: in each case, we see that there is a way to relabel the elements of R using the elements of S in a way that preserves the ring structure. ◦ The desired relabeling is a function ' : R ! S with the property that ' is a bijection (so that each element of R is labeled with a unique element of S) and that ' respects the ring operations. ◦ Explicitly, we require '(r1 + r2) = '(r1) + '(r2) and '(r1 · r2) = '(r1) · '(r2) for all elements r1 and r2 in R. • Denition: Let R and S be rings. A ring isomorphism ' from R to S is a bijective1 function ' : R ! S such that '(r1 + r2) = '(r1) + '(r2) and '(r1 · r2) = '(r1) · '(r2) for all elements r1 and r2 in R. 1Recall that a function ' : R ! S is injective (one-to-one) if '(x) = '(y) implies x = y, and ' is surjective (onto) if for every s 2 S there exists an r 2 R with '(r) = s. A bijective function is one that is both injective and surjective. Equivalently, ' is a bijection if it possesses a two-sided inverse function '−1 : S ! R with '('−1(s)) = s and '−1('(r)) = r for every r 2 R and s 2 S. 2 ◦ We remark here that in both of the conditions '(r1 + r2) = '(r1) + '(r2) and '(r1 · r2) = '(r1) · '(r2), the operations on the left are performed in R while the operations on the right are performed in S. ◦ Note: Isomorphisms arise in a variety of contexts (e.g., isomorphisms of vector spaces, isomorphisms of groups, etc.), and in some cases the rings we are considering may carry additional structure. We will simply say isomorphism rather than explicitly specifying ring isomorphism each time, unless there is a particular reason to do otherwise. • Example: For R = Z=6Z and S = (Z=2Z)×(Z=3Z), the map ' : R ! S dened via '(n) = (n mod 2; n mod 3) is an isomorphism. ◦ Note that reducing a residue class in Z=6Z modulo 2 or modulo 3 is well-dened, since 2 and 3 both divide 6, so ' is well-dened. ◦ It is then easy to see that ' is a bijection, and that '(r1 +r2) = '(r1)+'(r2) and '(r1 ·r2) = '(r1)·'(r2) for any residue classes . (It is also possible to compare the addition and multiplication tables r1; r2 2 Z=6Z as we did above.) ◦ We therefore conclude that ' is an isomorphism. a b • Example: For S = 2 M ( ): a; b 2 , show that the map ' : ! S dened via '(a+bi) = −b a 2×2 R R C a b is an isomorphism. −b a ◦ First, we see that ' is a bijection since it has a two-sided inverse; namely, the map '−1 : S ! C dened a b by '−1 = a + bi. −b a ◦ Furthermore, if z = a + bi and w = c + di, then a + c b + d a b c d '(z + w) = '((a + c) + (b + d)i) = = + = '(z) + '(w) −(b + d) a + c −b a −d c and also ac − bd ad + bc a b c d '(zw) = '((ac − bd) + (ad + bc)i) = = · = '(z) · '(w): −(ad + bc) ac − bd −b a −d c ◦ Thus, ' satises all the requirements, so it is an isomorphism. • Denition: If there is an isomorphism ' : R ! S, we say R and S are isomorphic, and write R =∼ S. ◦ Intuitively, isomorphic rings share the same structure, except that the elements and operations may be labeled dierently. • Proposition (Properties of Isomorphisms): If R; S; T are any rings, the following hold: 1. The identity map I : R ! R dened by I(r) = r for all r 2 R is an isomorphism from R to R.
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